Simple Harmonic Motion, platform and box

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Homework Help Overview

The problem involves a platform undergoing simple harmonic motion (SHM) with a specified amplitude, and a block resting on top of the platform. The question seeks to determine the shortest period of motion required for the block to remain in constant contact with the platform.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to use the maximum acceleration formula in relation to gravity to find the period of motion. Some participants question the correctness of the formulas being applied, particularly regarding the units and definitions of acceleration.

Discussion Status

Participants are actively discussing the validity of the formulas used, with one suggesting a correction to the maximum acceleration equation. There is no clear consensus yet, as the original poster expresses confusion over the feedback received.

Contextual Notes

There appears to be uncertainty regarding the correct application of formulas and the units involved in the calculations. The original poster also expresses frustration about the potential for misunderstanding the problem setup.

sheepcountme
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Homework Statement



A platform moves up and down in SHM, with amplitude 0.035 m. Resting on top of the platform is a block of wood. What is the shortest period of motion for the platform so that the block will remain in constant contact with it?


Homework Equations



a(t)=-A[tex]\omega[/tex]^2cos([tex]\omega[/tex]t+phase constant)
amax=A[tex]\omega[/tex]

The Attempt at a Solution



I didn't see how I could possibly use the first one with so many unknowns so I used amax=A[tex]\omega[/tex] and set amax=9.8 figuring that the amount couldn't be more than gravity otherwise the block and platform would separate (maybe I'm wrong in this).
And so I set 9.8=.035[tex]\omega[/tex] and solved (getting 280). The I used [tex]\omega[/tex]=2[tex]\pi[/tex]f and solving for f (getting 140/[tex]\pi[/tex]) and then using f=1/T and solved for T getting .0224 which was incorrect.
 
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Check your formula for amax, it is not correct.

ehild
 
Oh, it came out funny. It's supposed to be amax=A x omega. Is this still incorrect? If it is, I really need to throw a fit about a refund.
 
No, see the previous line. a must be m/s2, yours is m/s.

ehild
 

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